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Chapter 5: Problem 12

How many kcal will it take to heat 1 liter of water (e.g., in a pot) from room temperature \(\left(20^{\circ} \mathrm{C}\right)\) to boiling \(\left(100^{\circ} \mathrm{C}\right) ?\) How many Joules is this?

Short Answer

Expert verified
Answer: It takes 80 kcal or 334720 Joules to heat 1 liter of water from room temperature to boiling.

Step by step solution

01

Identify the given information

We are given: - Initial temperature: \(20^{\circ} \mathrm{C}\) - Final temperature: \(100^{\circ} \mathrm{C}\) - Volume of water: 1 liter = 1000 g (since the density of water is 1 g/mL) - Specific heat capacity of water: \(4.18 \frac{\mathrm{kJ}}{\mathrm{kg\cdot K}} = 1.00 \frac{\mathrm{kcal}}{\mathrm{kg\cdot K}}\)
02

Calculate the temperature difference

Delta T, or the change in temperature, is the difference between the final temperature and the initial temperature: \(\Delta T = (100 - 20)^{\circ} \mathrm{C} = 80^{\circ} \mathrm{C}\)
03

Calculate the energy required in kcal

To find the energy required to heat the water, use the formula: \(Q = mc\Delta T\) Where - \(Q\) is the energy required - \(m\) is the mass of the substance (1 kg, since 1000 g = 1 kg) - \(c\) is the specific heat capacity (1.00 \(\frac{\mathrm{kcal}}{\mathrm{kg\cdot K}}\)) - \(\Delta T\) is the change in temperature (80 \(^{\circ}\mathrm{C}\)) Plug in the values: \(Q = (1\,\mathrm{kg})\times(1.00\,\frac{\mathrm{kcal}}{\mathrm{kg\cdot K}})\times(80\,\mathrm{K})\) \(Q = 80\,\mathrm{kcal}\) It takes 80 kcal to heat 1 liter of water from room temperature to boiling.
04

Convert kcal to Joules

To convert 80 kcal to Joules, use the conversion factor: 1 kcal equals to 4184 Joules \(80\,\mathrm{kcal} \times \frac{4184\,\mathrm{J}}{1\,\mathrm{kcal}} = 334720\,\mathrm{J}\) It takes 334720 Joules to heat 1 liter of water from room temperature to boiling.

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